Help please !!!!.......
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Check the picture below.
so the focus point is at -2, 4 and the directrix is at y = 6, now, keeping in mind that the vertex is half-way between those two fellows, from 4 to 6, it'd be the y-coordinate of 5, and therefore, the vertex is at -2,5, as you see there in the picture, and the parabola looks like so. Since the parabola is a vertical one, the squared variable is the "x".
notice the distance "p", is just 1 unit, however, since the parabola is opening downwards, "p" is negative, and thus -1.
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2 \\\\\\ -4(y-5)=(x+2)^2\implies y-5=-\cfrac{1}{4}(x+2)^2 \\\\\\ y=-\cfrac{1}{4}(x+2)^2+5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A4p%28x-%20h%29%3D%28y-%20k%29%5E2%0A%5C%5C%5C%5C%0A%5Cboxed%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%0A%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D-2%5C%5C%0Ak%3D5%5C%5C%0Ap%3D-1%0A%5Cend%7Bcases%7D%5Cimplies%204%28-1%29%28y-5%29%3D%5Bx-%28-2%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A-4%28y-5%29%3D%28x%2B2%29%5E2%5Cimplies%20y-5%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%2B5)
so the focus point is at -2, 4 and the directrix is at y = 6, now, keeping in mind that the vertex is half-way between those two fellows, from 4 to 6, it'd be the y-coordinate of 5, and therefore, the vertex is at -2,5, as you see there in the picture, and the parabola looks like so. Since the parabola is a vertical one, the squared variable is the "x".
notice the distance "p", is just 1 unit, however, since the parabola is opening downwards, "p" is negative, and thus -1.